How do you solve $log_2(x+3)+log_2(x-3) = 4$ ?

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Answer 1 · 2018-04-05T22:29:46

Let's use one of the log laws, which says that $log_b(a) + log_b(c)$ can be rewritten as $log_b(a*c)$

$log_2(x+3)+log)2(x-3) = 4$

$log_2((x+3)(x-3)) = 4$

$log_2(x^2-9) = 4$

We can write a log , such as $log_b(a) = c$ to be $b^c = a$ (note, the base remains the same)

$2^4 = x^2 - 9$

$16 = x^2 - 9$

$25 = x^2$

$x = 5$

How do you solve log_2(x+3)+log_2(x-3) = 4log_2(x+3)+log_2(x-3) = 4?